$\overline{AB}$ = $3\sqrt{10}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $3\sqrt{10}$ $?$ $ \sin( \angle ABC ) = \frac{3\sqrt{10} }{10}, \cos( \angle ABC ) = \frac{ \sqrt{10}}{10}, \tan( \angle ABC ) = 3$
Solution: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is opposite to $\angle ABC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle ABC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{3\sqrt{10}} $ $ \overline{AC}=3\sqrt{10} \cdot \sin( \angle ABC ) = 3\sqrt{10} \cdot \frac{3\sqrt{10} }{10} = 9$